Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 15

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Weekly Quiz Competition

Question 1
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If $$P$$ is a point on the line passing through the point $$A$$ with position vector $$2\overline{i}+\overline{j}-3\overline{k}$$ and parallel to $$\overline{i}+2\overline{j}+\overline{k}$$ such that $$AP=2\sqrt{6}$$ then the position vector of $$P$$ is

A
$$4\overline{i}+5\overline{j}+\overline{k}$$

B
$$3\overline{j}+5\overline{k}$$

C
$$4\overline{i}+5\overline{j}-\overline{k}$$

D
$$3\overline{j}-4\overline{k}$$

Solution

$$P\cdot V\ of\ P= (2+t)\hat{i}+(1+2t)\hat{j}+(-3+t)\hat{k}$$
$$P\cdot V\ of\ A= (2,1,-3)$$
$$\left | \overrightarrow{AP} \right |= \sqrt{t^{2}+4t^{2}+t^{2}}$$
$$2\ \sqrt{6}= t\sqrt{6}$$
$$\Rightarrow t = 2$$
$$P= 4\hat{i}+5\hat{j}-\hat{k}$$
Question 2
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Cosine of the angle between two diagonals of a cube is equal to

A
$$\displaystyle \frac{2}{\sqrt 6}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{2}$$

D
None of these

Solution

Let 0, one vertex of a cube, be the origin and three edges through O be the coordinate axes. The four diagonals are Op, AA', BB' and CC'. Let a be the length of each edge. Then the coordinates of P, A, A' are (a, a, a), (a, 0, 0), (0, a, a).
The direction ratios of OP are a, a, a.
The direction cosines of OP are
$$\displaystyle \frac{a}{a \sqrt 3}, \frac{a}{a \sqrt 3}, \frac{a}{a \sqrt 3}$$ i.e., $$\displaystyle \frac{1}{\sqrt 3} , \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}$$.
Similarly direction cosines of AA'are
$$\displaystyle \left ( - \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3} \right )$$.
Let $$\theta$$ be the angle between the diagonals OP and AA'.
$$cos \theta = \displaystyle \frac{1}{\sqrt 3} \left ( - \frac{1}{\sqrt 3} \right ) + \frac{1}{\sqrt 3} \left ( \frac{1}{\sqrt 3} \right ) + \frac{1}{\sqrt 3} \left (\frac{1}{\sqrt 3} \right )$$
$$cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 = - \displaystyle \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1}{3}$$
$$\therefore \displaystyle \theta = cos^{-1} \left ( \frac{1}{3} \right )$$.
Question 3
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Directions For Questions

Read the following write up carefully and answer the following questions
There are two planes $$\displaystyle P_{1}:x-2y+3z=15$$ and plane $$\displaystyle P_{2}:x-2y+3z=30$$ Lines $$\displaystyle \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-6}{1},\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-6}{1}\: and\: \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-6}{-2}$$ passes through point P on the plane $$\displaystyle P_{1}$$ and intersects plane $$\displaystyle P_{2}$$ at A, B, C respectively Then

...view full instructions

Area of $$\displaystyle \Delta ABC$$ is

A
$$45$$ squares units

B
$$55$$ squares units

C
$$65$$ squares units

D
none of these

Solution

Line PA : $$\displaystyle \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-6}{1}$$
Line PB : $$\displaystyle \frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-6}{1}$$
Line PC : $$\displaystyle \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-6}{-2}$$
Then $$\displaystyle A\left ( \frac{7}{2},-\frac{1}{2},\frac{17}{2} \right )$$
$$\displaystyle B\left ( \frac{17}{2},-13,-\frac{3}{2} \right )$$
$$\displaystyle C\left ( -14,\frac{19}{2},21 \right )$$
Hence area of $$\displaystyle \Delta ABC=\frac{225\sqrt{14}}{8}$$, volume of tetrahedron PABC = $$\displaystyle \frac{1125}{8}$$ cubic units
Question 4
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A vector is equally inclined to the $$x$$-axis, $$y$$-axis and $$z$$-axis respectively, its direction cosines are

A
$$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$

B
$$< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$

C
$$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$ or $$< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$

D
None of these

Solution

Let the vector $$v$$ make an angle $$\alpha$$ with each of the three axes, then direction cosine of $$v$$ are
$$<\cos\,\alpha, \cos\,\alpha, \cos\,\alpha>$$
$$\Rightarrow \cos^2\alpha =\dfrac {1}{3}$$
$$\Rightarrow \cos\,\alpha=\pm \dfrac{1}{\sqrt{3}}$$
Hence, direction cosine of $$v$$ are
$$<\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$ or
$$<-\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$
Question 5
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Find vector equation of line passing through $$(1,4,1),(1,2,2)$$.

A
$$\overline r=\overline i+(4-2\lambda)\overline j+(1+\lambda)\overline k$$

B
$$\overline r=\overline i+(4+2\lambda)\overline j+(1+\lambda)\overline k$$

C
$$\overline r=\overline i-(4-2\lambda)\overline j+(1+\lambda)\overline k$$

D
$$\overline r=\overline i+(4-2\lambda)\overline j+(1-\lambda)\overline k$$

Solution

Given points
$$A(1,4,1)$$ and $$B(1,2,2)$$
position vector of point A
$$\vec{a}=\hat{i}+4\hat{j}+\hat{k}$$
position vector of point B
$$\vec{b}=\hat{i}+2\hat{j}+2\hat{k}$$
normal vector of line be
$$\vec{n}=\vec{b}-\vec{a}$$
$$\vec{n}=\hat{i}+2\hat{j}+2\hat{k}-(\hat{i}+4\hat{j}+\hat{k})$$
$$\vec{n}=\hat{i}+2\hat{j}+2\hat{k}-\hat{i}-4\hat{j}-\hat{k}$$
$$\vec{n}=-2\hat{j}+\hat{k}$$
vector eq of line
$$\vec{r}=\vec{a}+\lambda(\vec{n})$$
$$\vec{r}=\hat{i}+4\hat{j}+\hat{k}+\lambda(-2\hat{j}+\hat{k})$$
$$\vec{r}=\hat{i}+4\hat{j}+\hat{k}-2\lambda\hat{j}+\lambda\hat{k})$$
$$\vec{r}=\hat{i}+(4-2\lambda)\hat{j}+(1+\lambda)\hat{k}$$
Question 6
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What are the DR's of vector parallel to $$\left( 2,-1,1 \right) $$ and $$\left( 3,4,-1 \right) $$?

A
$$\left( 1,5,-2 \right) $$

B
$$\left( -2,-5,2 \right) $$

C
$$\left( -1,5,2 \right) $$

D
$$\left( -1,-5,-2 \right) $$

Solution

Required DR's are $$\left( 3-2,4+1,-1-1 \right) $$ ie, $$\left( 1,5,-2 \right) $$
Question 7
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The direction cosines of the vectors $$2\vec {i} + \vec {j} - 2\vec {k}$$ is equal to

A
$$\dfrac {2}{3}, \dfrac {1}{3}, -\dfrac {2}{3}$$

B
$$\dfrac {2}{3}, \dfrac {1}{3}, \dfrac {2}{3}$$

C
$$\dfrac {1}{3}, \dfrac {2}{3}, -\dfrac {2}{3}$$

D
$$\dfrac {2}{3}, \dfrac {2}{3}, \dfrac {1}{3}$$

Solution

We know, $$\dfrac{a_{1}}{\left|a\right|},\dfrac{a_{2}}{\left|a\right|},\dfrac{a_{3}}{\left|a\right|}$$ are the direction cosines of $$a_{1}i+a_{2}j+a_{3}k$$
Given vector is $$2\hat i+\hat j-2\hat k$$
$$|a|=\sqrt {2^2+1^2+2^2}=\sqrt {9}=3$$
Thus their direction cosines are $$\dfrac {2}{3},\dfrac {1}{3},- \dfrac {2}{3}$$
So, option A is correct.
Question 8
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The Cartesian equation of the plane passing through the point $$(3, -2, -1)$$ and parallel to the vectors $$\overline {b} = \overline {i} - 2\overline {j} + 4\overline {k}$$ and $$\overline {c} = 3\overline {i} + 2\overline {j} - 5\overline {k}$$ is

A
$$2x - 17y - 8z + 63 = 0$$

B
$$3x + 17y + 8z - 36 = 0$$

C
$$2x + 17y + 8z + 36 = 0$$

D
$$3x - 16y + 8z - 63 = 0$$

Solution

$$\begin{vmatrix} x - x_{1}& y - y_{1} & z - z_{1}\\ a_{1} & b_{1} &c_{1} \\ a_{2} & b_{2} & c_{2} \end{vmatrix} = 0$$

$$\begin{vmatrix} x - 3 & y + 2 & z + 1\\ 1 & -2 & 4 \\ 3 & 2 & -5 \end{vmatrix} = 0$$

$$2x+ 17y +8z +36= 0$$
Question 9
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A straight line is equally inclined to all the three coordinate axes. Then an angle made by the line with the y-axis is

A
$$\cos^{1}\left (\dfrac {1}{3}\right )$$

B
$$\cos^{1}\left (\dfrac {1}{\sqrt {3}}\right )$$

C
$$\cos^{1}\left (\dfrac {2}{\sqrt {3}}\right )$$

D
$$\dfrac {\pi}{4}$$

Solution

Let $$l, m, n$$ be the directional cosines of the line.
We know that $$l^{2}+m^{2}+n^{2} = 1$$
Since the line is equally inclined to all the axes, $$l = m = n$$

Solving the two above equations we get, $$l = m = n = \frac{1}{\sqrt{3}}$$
Hence the angle made be the line with any axis is,
$$cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$
Question 10
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If the foot of the perpendicular from $$(0, 0, 0)$$ to a plane is $$(1, 2, 3)$$, then the equation of the plane is

A
$$2x + y + 3z = 14$$

B
$$x + 2y + 3z = 14$$

C
$$x + 2y + 3z + 14 = 0$$

D
$$x + 2y - 3z = 14$$

Solution

The normal of the plane passes through $$(1, 2, 3) and (0, 0, 0).$$
Hence the normal vector is given by $$1\hat{i}+2\hat{j}+3\hat{k}$$
The equation of the plane is of the form $$x+2y+3z = c$$, where c is a constant.
Since the plane passes through the point $$(1, 2, 3)$$,
$$1(1)+2(2)+3(3) = c$$
$$c = 14$$

Hence the equation of the plane is $$x + 2y+3z = 14$$

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 15

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Three Dimensional Geometry Test - 15

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SHARING IS CARING

Question 1

$$4\overline{i}+5\overline{j}+\overline{k}$$

$$3\overline{j}+5\overline{k}$$

$$4\overline{i}+5\overline{j}-\overline{k}$$

$$3\overline{j}-4\overline{k}$$

Solution

Question 2

$$\displaystyle \frac{2}{\sqrt 6}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{1}{2}$$

None of these

Solution

Question 3

$$45$$ squares units

$$55$$ squares units

$$65$$ squares units

none of these

Solution

Question 4

$$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$

$$< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$

$$< \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}>$$ or $$< -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}>$$

None of these

Solution

Question 5

$$\overline r=\overline i+(4-2\lambda)\overline j+(1+\lambda)\overline k$$

$$\overline r=\overline i+(4+2\lambda)\overline j+(1+\lambda)\overline k$$

$$\overline r=\overline i-(4-2\lambda)\overline j+(1+\lambda)\overline k$$

$$\overline r=\overline i+(4-2\lambda)\overline j+(1-\lambda)\overline k$$

Solution

Question 6

$$\left( 1,5,-2 \right) $$

$$\left( -2,-5,2 \right) $$

$$\left( -1,5,2 \right) $$

$$\left( -1,-5,-2 \right) $$

Solution

Question 7

$$\dfrac {2}{3}, \dfrac {1}{3}, -\dfrac {2}{3}$$

$$\dfrac {2}{3}, \dfrac {1}{3}, \dfrac {2}{3}$$

$$\dfrac {1}{3}, \dfrac {2}{3}, -\dfrac {2}{3}$$

$$\dfrac {2}{3}, \dfrac {2}{3}, \dfrac {1}{3}$$

Solution

Question 8

$$2x - 17y - 8z + 63 = 0$$

$$3x + 17y + 8z - 36 = 0$$

$$2x + 17y + 8z + 36 = 0$$

$$3x - 16y + 8z - 63 = 0$$

Solution

Question 9

$$\cos^{1}\left (\dfrac {1}{3}\right )$$

$$\cos^{1}\left (\dfrac {1}{\sqrt {3}}\right )$$

$$\cos^{1}\left (\dfrac {2}{\sqrt {3}}\right )$$

$$\dfrac {\pi}{4}$$

Solution

Question 10

$$2x + y + 3z = 14$$

$$x + 2y + 3z = 14$$

$$x + 2y + 3z + 14 = 0$$

$$x + 2y - 3z = 14$$

Solution

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